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# NCERT Exemplar- Application of Derivatives (Part - 2) Notes | EduRev

## JEE : NCERT Exemplar- Application of Derivatives (Part - 2) Notes | EduRev

The document NCERT Exemplar- Application of Derivatives (Part - 2) Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Objective Type Questions
Q.35. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is:
(a) 10 cm2/s
(b) âˆš3 cm2/s
(c) 10 âˆš3 cm2/s
(d) 10/3 cm2/s
Ans. (c)
Solution.
Let the length of each side of the given equilateral triangle be x cm.
âˆ´
Area of equilateral triangle
âˆ´
Hence, the rate of increasing of area = 10 âˆš3 cm2/sec .
Hence, the correct option is (c).

Q.36. A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is:

Ans. (b)
Solution.
Length of ladder = 5 m
Let AB = y m and BC = x m
âˆ´ In right Î”ABC,
AB2 + BC2 = AC2

â‡’ x2 + y2 = (5)2 â‡’ x2 + y2 = 25
Differentiating both sides w.r.t x, we have

â‡’
â‡’[âˆµ x = 2m]
â‡’
â‡’x (- 0.1) = 0
â‡’
â‡’
Differentiating both sides w.r.t. t, we get

â‡’
[(â€“) sign shows the decrease of change of angle]
Hence, the correct option is (b).

Q.37. The curve y =x1/5 has at (0, 0)
(a) a vertical tangent (parallel to y-axis)
(b) a horizontal tangent (parallel to x-axis)
(c) an oblique tangent
(d) no tangent
Ans. (a)
Solution.
Equation of curve is y = x1/5
Differentiating w.r.t. x, we get
(at x = 0)

âˆ´ The tangent is parallel to y-axis.
Hence, the correct option is (a).

Q.38. The equation of normal to the curve 3x2 â€“ y2 = 8 which is parallel to the line x + 3y = 8 is
(a) 3x â€“ y = 8
(b) 3x + y + 8 = 0
(c) x + 3y Â± 8 = 0
(d) x + 3y = 0
Ans. (c)
Solution.
Given equation of the curve is 3x2 â€“ y2 = 8 ...(i)
Differentiating both sides w.r.t. x, we get

is the slope of the tangent
âˆ´ Slope of the normal =
Now x + 3y = 8 is parallel to the normal
Differentiating both sides w.r.t. x, we have

âˆ´
Putting y = x in eq. (i) we get
3x2 â€“ x2 = 8 â‡’ 2x2 = 8 â‡’ x2 = 4
âˆ´ x = Â± 2 and y = Â± 2
So the points are (2, 2) and (â€“ 2, â€“ 2).
Equation of normal to the given curve at (2, 2) is

â‡’ 3y â€“ 6 = â€“ x + 2 â‡’ x + 3y â€“ 8 = 0
Equation of normal at (â€“ 2, â€“ 2) is

â‡’ 3y + 6 = â€“ x â€“ 2 â‡’ x + 3y + 8 = 0
âˆ´ The equations of the normals to the curve are
x + 3y Â± 8 = 0
Hence, the correct option is (c).

Q.39. If the curve ay + x2 = 7  and x3 = y, cut orthogonally at (1, 1), then the value of a is:

(a) 1
(b) 0
(c) â€“ 6
(d) 6
Ans. (d)
Solution.
Equation of the given curves are ay + x2 = 7 ...(i)
and x3 = y ...(ii)
Differentiating eq. (i) w.r.t. x, we have

âˆ´
Now differentiating eq. (ii) w.r.t. x, we get

The two curves are said to be orthogonal if the angle between the tangents at the point of intersection is 90Â°.

(1, 1) is the point of intersection of two curves.
âˆ´ 6(1)3 = a
So a = 6
Hence, the correct option is (d).

Q.40. If y = xâ€“ 10 and if x changes from 2 to 1.99, what is the change in y
(a) 0.32
(b) 0.032
(c) 5.68
(d) 5.968
Ans. (a)
Solution.

Given that y = x4 â€“ 10

Î”x = 2.00 â€“ 1.99 = 0.01
âˆ´

= 4 x (2)3 x 0.01 = 32 x 0.01 = 0.32
Hence, the correct option is (a).

Q.41. The equation of tangent to the curve y (1 + x2) = 2 â€“ x, where it crosses x-axis is:
(a) x + 5y = 2
(b) x â€“ 5y = 2
(c) 5x â€“ y = 2
(d) 5x + y = 2
Ans. (a)
Solution.

Given that y(1 + x2) = 2 â€“ x ...(i)
If it cuts x-axis, then y-coordinate is 0.
âˆ´ 0(1 + x2) = 2 â€“ x â‡’ x = 2
Put x = 2 in equation (i)
y(1 + 4) = 2 â€“ 2 â‡’ y(5) = 0 â‡’ y = 0
Point of contact = (2, 0)
Differentiating eq. (i) w.r.t. x, we have

Equation of tangent is y â€“ 0 =
â‡’ 5y = â€“ x + 2 â‡’ x + 5y = 2
Hence, the correct option is (a).

Q.42. The points at which the tangents to the curve y = x3 â€“ 12x + 18 are parallel to x-axis are:
(A) (2, â€“2), (â€“2, â€“34)
(B) (2, 34), (â€“2, 0)
(C) (0, 34), (â€“2, 0)
(D) (2, 2), (â€“2, 34)
Ans. (d)
Solution.

Given that y = x3 â€“ 12x + 18
Differentiating both sides w.r.t. x, we have
â‡’
Since the tangents are parallel to x-axis, then
âˆ´ 3x2 â€“ 12 = 0 â‡’ x = Â± 2
âˆ´ yx=2 = (2)3 â€“ 12(2) + 18 = 8 â€“ 24 + 18 = 2
yx=â€“2  = (â€“ 2)3 â€“ 12(â€“ 2) + 18 = â€“ 8 + 24 + 18 = 34
âˆ´ Points are (2, 2) and (â€“ 2, 34)
Hence, the correct option is (d).

Q.43. The tangent to the curve y = e2x at the point (0, 1) meets x-axis at:
(a) (0, 1)
(b)
(c) (2, 0)
(d) (0, 2)
Ans. (b)
Solution.

Equation of the curve is y = e2x
Slope of the tangent
âˆ´ Equation of tangent to the curve at (0, 1) is
y â€“1 = 2(x â€“ 0)
â‡’ y â€“ 1 = 2x â‡’ y â€“ 2x = 1
Since the tangent meets x-axis where y = 0
âˆ´ 0 â€“ 2x = 1 â‡’
So the point is
Hence, the correct option is (b).

Q.44. The slope of tangent to the curve x = t2 + 3t â€“ 8, y = 2t2 â€“ 2t â€“ 5 at the point (2, â€“1) is:

(a) 22/ 7
(b) 6/7
(c) - 6/7
(d) â€“ 6
Ans. (b)
Solution.

The given curve is x = t2 + 3t â€“ 8 and y = 2t2 â€“ 2t â€“ 5

âˆ´
Now (2, â€“ 1) lies on the curve
â‡’ 2 = t2 + 3t â€“ 8 â‡’ t2 + 3t â€“ 10 = 0
â‡’ t2 + 5t â€“2t â€“ 10 = 0
â‡’ t(t + 5) â€“ 2(t + 5) = 0
â‡’ (t + 5) (t â€“ 2) = 0
âˆ´ t = 2, t = â€“ 5 and â€“ 1 = 2t2 â€“ 2t â€“ 5
â‡’ 2t2 â€“ 2t â€“ 4 = 0
â‡’ t2 â€“ t â€“ 2 = 0 â‡’ t2 â€“ 2t + t â€“ 2 = 0
â‡’ t(t â€“ 2) + 1 (t â€“ 2) = 0 â‡’ (t + 1) (t - 2) = 0
â‡’ t = â€“ 1 and t = 2
So t = 2 is common value
âˆ´
Hence, the correct option is (b).

Q.45. The two curves xâ€“ 3xy2 + 2 = 0 and 3x2y â€“ y3 â€“ 2 = 0 intersect at an angle of
(a) Ï€/4
(b) Ï€/3
(c) Ï€/2
(d) Ï€/6
Ans. (c)
Solution.

The given curves are x3 â€“ 3xy2 + 2 = 0 ...(i)
and 3x2y â€“ y3 â€“ 2 = 0 ...(ii)
Differentiating eq. (i) w.r.t. x, we get

â‡’
âˆ´
So slope of the curve
Differentiating eq. (ii) w.r.t. x, we get

âˆ´
So the slope of the curve
Now
So the angle between the curves is
Hence, the correct option is (c).

Q.46. The interval on which the function f (x) = 2x3 + 9x2 + 12x â€“ 1 is decreasing is:
(a) [â€“1, âˆž )
(b) [â€“2, â€“1]
(c) (â€“ âˆž , â€“2]
(d) [â€“1, 1]
Ans. (b)
Solution.

The given function is f (x) = 2x3 + 9x2 + 12x - 1
f  â€²(x) = 6x2 + 18x + 12
For increasing and decreasing f  â€²(x) = 0
âˆ´ 6x2 + 18x + 12 = 0
â‡’ x2 + 3x + 2 = 0 â‡’ x2 + 2x + x + 2 = 0
â‡’ x(x + 2) + 1(x + 2) = 0 â‡’ (x + 2) (x + 1) = 0
â‡’ x = â€“ 2, x = â€“ 1
The possible intervals are (â€“âˆž, - 2), ( - 2, - 1), ( - 1, âˆž)
Now f â€²(x) = (x + 2) (x + 1)
â‡’ f â€²(x)( - âˆž, - 2) = (-) (-) = (+) increasing
â‡’ f â€²(x)(- 2, - 1) = (+) (-) = (-) decreasing
â‡’ f â€²(x)(- 1, âˆž) = (+) (+) = (+) increasing
Hence, the correct option is (b).

Q.47. Let the f : R â†’ R be defined by f (x) = 2x + cosx, then f :
(a) has a minimum at x = Ï€
(b) has a maximum, at x = 0
(c) is a decreasing function
(d) is an increasing function
Ans. (d)
Solution.

Given that f (x) = 2x + cos x
f â€²(x) = 2 - sin x
Since f â€²(x) > 0 âˆ€ x
So f (x) is a n increasing function.
Hence, the correct option is (d).

Q.48. y = x (x â€“ 3)2 decreases for the values of x given by :
(a) 1 < x < 3
(b) x < 0
(c) x > 0
(d) 0 < x < 3/2
Ans. (a)
Solution.

Here y  = x(x â€“ 3)2

For increasing and decreasing
âˆ´ 2x(x â€“ 3) + (x â€“ 3)2 = 0 â‡’ (x - 3) (2x + x - 3) = 0
â‡’ (x â€“ 3) (3x â€“ 3) = 0 â‡’ 3(x - 3) (x - 1) = 0
âˆ´ x = 1, 3
âˆ´ Possible intervals are (â€“ âˆž, 1), (1, 3), (3, âˆž)
dy/dx = (x - 3) (x - 1)
For (- âˆž , 1) =(-) (-) = (+) increasing
For (1, 3) =(-) (+) = (-) decreasing
For (3, âˆž) =(+) (+) = (+) increasing
So the function decreases in (1, 3) or 1 < x < 3
Hence, the correct option is (a).

Q.49. The function f (x) = 4 sin3x â€“ 6 sin2x + 12 sinx + 100 is strictly
(a) increasing in
(b) decreasing in
(c) decreasing in
(d) decreasing in
Ans. (b)
Solution.

Here,
f (x) = 4 sin3 x â€“ 6 sin2 x + 12 sin x + 100
f â€²(x) = 12 sin2 x Ã— cos x - 12 sin x cos x + 12 cos x
= 12 cos x [sin2 x - sin x + 1]
= 12 cos x [sin2 x + (1 - sin x)]
âˆ´ 1 - sin x â‰¥ 0 and sin2 x â‰¥ 0
âˆ´ sin2 x + 1 - sin x â‰¥ 0 (when cos x > 0)
Hence, f â€²(x) > 0 , when cos x > 0 i.e.,
So, f(x) is in creasing whereand f â€²(x) < 0
when cos x < 0 i.e.
Hence, (x) is decreasing when

So f(x) is decreasing in
Hence, the correct option is (b).

Q.50. Which of the following functions is decreasing on
(a) sin 2x
(b) tan x
(c) cos x
(d) cos 3x
Ans. (c)
Solution.

Here, Let f (x) = cos x;  So, f â€²(x) = - sin x
f â€²(x) < 0 in
So f (x) = cos x is decreasing in
Hence, the correct option is (c).

Q.51. The function f (x) = tanx â€“ x
(a) always increases
(b) always decreases
(c) never increases
(d) sometimes increases and sometimes decreases.
Ans. (a)
Solution.

Here, f(x) = tan x â€“ x So, f â€²(x) = sec2 x - 1
f â€²(x) > 0 âˆ€ x âˆˆ R
So f (x) is always increasing
Hence, the correct option is (a).

Q.52. If x is real, the minimum value of x2 â€“ 8x + 17 is
(a) â€“ 1
(b) 0
(c) 1
(d) 2
Ans. (c)
Solution.

Let f (x) = x2 â€“ 8x + 17
f â€²(x) = 2x - 8
For local maxima and local minima, f â€²(x) = 0
âˆ´ 2x â€“ 8 = 0 â‡’ x = 4
So, x = 4 is  the point of local maxima and local minima.
f â€³(x) = 2 > 0 minima at x = 4
âˆ´ f(x)x=4 = (4)2 â€“ 8(4) + 17
= 16 â€“ 32 + 17 = 33 â€“ 32 = 1
So the minimum value of the function is 1
Hence, the correct option is (c).

Q.53. The smallest value of the polynomial x3 â€“ 18x2 + 96x in [0, 9] is
(a) 126
(b) 0
(c) 135
(d) 160
Ans. (b)
Solution.

Let f(x) = x3 â€“ 18x2 + 96x; So, f â€²(x) = 3x2 - 36x + 96
For local maxima and local minima f  â€²(x) = 0
âˆ´ 3x2 - 36x + 96 = 0
â‡’ x2 - 12x + 32 = 0 â‡’ x2 - 8x - 4x + 32 = 0
â‡’ x(x - 8) - 4(x - 8) = 0 â‡’ (x - 8) (x - 4) = 0
âˆ´ x = 8, 4 âˆˆ [0, 9]
So, x = 4, 8  are the points of local maxima and local minima.
Now we will calculate the absolute maxima or absolute
minima at x = 0, 4, 8, 9
âˆ´ f(x) = xâ€“ 18x2 + 96x
f(x)x=0 = 0 â€“ 0 + 0 = 0
f(x)x=4 = (4)3 â€“ 18(4)2 + 96(4)
= 64 â€“ 288 + 384 = 448 â€“ 288 = 160
f(x)x=8 = (8)3 â€“ 18(8)2 + 96(8)
= 512 â€“ 1152 + 768 = 1280 â€“ 1152 = 128
f(x)x=9 = (9)3 â€“ 18(9)2 + 96(9)
= 729 â€“ 1458 + 864 = 1593 â€“ 1458 = 135
So, the absolute minimum value of f is 0 at x = 0
Hence, the correct option is (b).

Q.54. The function f (x) = 2x3 â€“ 3x2 â€“ 12x + 4, has
(a) two points of local maximum
(b) two points of local minimum
(c) one maxima and one minima
(d) no maxima or minima
Ans. (c)
Solution.

We have f(x) = 2x3 â€“ 3x2 â€“ 12x + 4
f â€²(x) = 6x2 - 6x - 12
For local maxima and local minima f  â€²(x) = 0
âˆ´ 6x2 - 6x - 12 = 0
â‡’ x2 â€“ x â€“ 2 = 0 â‡’ x2 â€“ 2x + x â€“ 2 = 0
â‡’ x(x â€“ 2) + 1(x â€“ 2) = 0 â‡’ (x + 1) (x - 2) = 0
â‡’ x = -1, 2 are the points of local maxima and local minima
Now f â€³(x) = 12x - 6
f â€³(x)x = - 1 = 12(-1) - 6 = - 12 - 6 = - 18 < 0, maxima
f â€³(x)x = 2 = 12(2) - 6 = 24 - 6 = 18 > 0 minima
So, the function is maximum at x = -1 and minimum at x = 2
Hence, the correct option is (c).

Q.55. The maximum value of sin x . cos x is
(a) 1/4
(b) 1/2
(c) âˆš2
(d) 2âˆš2

Ans. (b)
Solution.

We have f (x) = sin x cos x
â‡’

â‡’ f â€²(x) = cos 2x
Now for local maxima and local minima f  â€²(x) = 0
âˆ´ cos 2x = 0

â‡’
âˆ´
f â€³(x) = - 2 sin 2x

So f (x) is maximum at
âˆ´ Maximum value of f(x) =
Hence, the correct option is (b).

Q.56. At, f (x) = 2 sin3x + 3 cos3x is:
(a) maximum
(b) minimum
(c) zero
(d) neither maximum nor minimum
Ans. (a)
Solution.

We have f (x) = 2 sin 3x + 3 cos 3x
f â€²(x) = 2 cos 3x Ã— 3 - 3 sin 3xÃ—3 = 6 cos 3x - 9 sin 3x
f â€³(x) = - 6 sin 3x Ã— 3 - 9 cos 3x Ã— 3
= - 18 sin 3x - 27 cos 3x

=â€“ 18 < 0
maxima Maximum value of f(x) at

Hence, the correct option is (a).

Q.57. Maximum slope of the curve y = â€“x3 + 3x2 + 9x â€“ 27 is:
(a) 0
(b) 12
(c) 16
(d) 32
Ans. (b)
Solution.

Given that y = â€“ x3 + 3x+ 9x â€“ 27

âˆ´ Slope of the given curve,
m = â€“ 3x2 + 6x + 9

For local maxima and local minima,
âˆ´â€“ 6x + 6 = 0 â‡’ x = 1
Now
Maximum value of the slope at x = 1 is
mx = 1 = â€“ 3(1)2 + 6(1) + 9 = â€“ 3 + 6 + 9 = 12
Hence, the correct option is (b).

Q.58. f (x) = xx has a stationary point at
(a) x = e
(b)x = 1/e

(c) x = 1
(d) x =âˆše

Ans. (b)
Solution.

We have f (x) = xx
Taking log of both sides, we have
log f (x) = x log x
Differentiating both sides w.r.t. x, we get

â‡’ f â€²(x) = f (x) [1 + log x] = xx [1 + log x]
To find stationary point, f â€²(x) = 0
âˆ´ xx[1 + log x] = 0
xx â‰  0 âˆ´ 1 + log x = 0
â‡’ log x = â€“ 1  x = eâ€“1 â‡’
Hence, the correct option is (b).

Q.59. The maximum value of
(a) e
(b) ee
(c) e1/e
(d)
Ans. (c)
Solution.

Let
Taking log on both sides, we get

â‡’  log [f (x)] = x log x-1 â‡’  log [f (x)] = - [x log x]
Differentiating both sides w.r.t. x, we get

â‡’
For local maxima and local minima f â€²(x) = 0

âˆ´ 1 + log x = 0 â‡’ log x = â€“ 1 â‡’ x = eâ€“1
So,is the stationary point.
Now

âˆ´ Maximum value of the function at

Hence, the correct option is (c).

Fill in the blanks

Q.60. The curves y = 4x2 + 2x â€“ 8 and y = x3 â€“ x + 13 touch each other at the point_____.
Ans.
We have
y = 4x2 + 2x â€“ 8 ...(i)
and y = x3 â€“ x + 13 ...(ii)
Differentiating eq. (i) w.r.t. x, we have

[m is the slope of curve (i)]
Differentiating eq. (ii) w.r.t. x, we get
= 3x2 â€“ 1 â‡’ m2 = 3x2 â€“ 1
[m2 is the slope of curve (ii)]
If the two curves touch each other, then m1 = m2
âˆ´ 8x + 2 = 3x2 â€“ 1
â‡’ 3x2 â€“ 8x â€“ 3 = 0 â‡’ 3x2 - 9x + x - 3 = 0
â‡’ 3x(x - 3) + 1(x - 3) = 0 â‡’ (x - 3) (3x + 1) = 0
âˆ´
Putting x = 3 in eq. (i), we get
y = 4(3)2 + 2(3) â€“ 8 = 36 + 6 â€“ 8 = 34
So, the required point is (3, 34)
Now for x =

âˆ´ Other required point is
Hence, the required points are (3, 34) and

Q.61. The equation of normal to the curve y = tanx at (0, 0) is ________.
Ans.
We have y = tan x. So,
âˆ´ Slope of the normal =
at the point (0, 0) the slope = â€“ cos2 (0) = â€“1
So the equation of normal at (0, 0) is y â€“ 0 = â€“1(x â€“ 0)
â‡’ y = â€“ x â‡’ y + x =0
Hence, the required equation is y + x = 0.

Q.62. The values of a for which the function f (x) = sinx â€“ ax + b increases on R are ______.
Ans.
We have f (x) = sin x â€“ ax + b â‡’ f â€²(x) = cos x - a
For increasing the function f â€²(x) > 0
âˆ´ cos x - a > 0
Since cos x âˆˆ [ - 1, 1]
âˆ´ a < â€“1 â‡’ a âˆˆ ( - âˆž, - 1)
Hence, the value of a is ( - âˆž, - 1).

Q.63. The functiondecreases in the interval _______.
Ans.
We have f (x) =

For decreasing the function f â€²(x) < 0
âˆ´
âˆ´ x > Â± 1 â‡’ x âˆˆ (1, âˆž)
Hence, the required interval is (1, âˆž).

Q.64. The least value of the function f (x) =
is ______.
Ans.
Here,
For maximum and minimum value f â€²(x) = 0
âˆ´
Now
(âˆµ a, b > 0)
Hence,
minima So the least value of the function at x

Hence, least value is

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## Mathematics (Maths) Class 12

209 videos|222 docs|124 tests

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